Codeforces Round #160 (Div. 2)---A. Roma and Lucky Numbers

Roma and Lucky Numbers
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers.
Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Roma's got n positive integers. He wonders, how many of those integers have not more than k lucky digits? Help him, write the program that solves the problem.
Input
The first line contains two integers n, k (1 ≤ n, k ≤ 100). The second line contains n integers ai (1 ≤ ai ≤ 109) — the numbers that Roma has.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Sample test(s)
input

3 4
1 2 4

output

3

input

3 2
447 44 77

output

2

Note
In the first sample all numbers contain at most four lucky digits, so the answer is 3.
In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2.
제목의 대의: 뇌장애야, 처음에는 뜻밖에도 제목의 뜻을 이해하지 못했다.n개의 수를 제시하여 이 다수 중 몇 개가 k보다 많지 않은 행운의 숫자를 포함하는지 보여 줍니다. 그 중 4와 7은 행운의 숫자입니다.
문제풀이 사고방식: 매 수의 만족 여부를 직접 추정하면 된다.
AC 코드:

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff

int main()
{
    #ifdef sxk
        freopen("in.txt","r",stdin);
    #endif
    int n, k, t;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        int ans = 0;
        for(int i=0; i<n; i++){
            scanf("%d", &t);
                int cnt = 0;;
                while(t){
                    int x = t%10;
                    if(x==4 || x==7) cnt ++;
                    t /= 10;
                }
            if(cnt <= k) ans ++;
        }
        printf("%d
", ans);
    }
    return 0;
}